1. Discrete Maths Objective to introduce mathematical induction through examples 242-213, Semester 2, 2014-2015 8. Mathematical Induction 1

2. Overview 1. Motivation 2. Induction Defined 3.Maths Notation Reminder 4.Four Examples 5.More General Induction Proofs 6.Further Information 2

3. 1. Motivation Induction is used in mathematical proofs of recursive algorithms e.g. quicksort, binary search Induction is used to mathematically define recursive data structures e.g. lists, trees, graphs continued 3 Part 9

4. Induction is often used to derive mathematical estimates of program running time timings based on the size of input data e.g. time increases linearly with the number of data items processed timings based on the number of times a loop executes 4 Part 10

5. 2. Induction Defined Induction is used to solve problems such as: is S(n) correct/true for all n values? usually for all n >= 0 or all n >=1 Example: let S(n) be "n 2 + 1 > 0" is S(n) true for all n >= 1? continued S(n) can be much more complicated, such as a program that reads in a n value. S(n) can be much more complicated, such as a program that reads in a n value. 5

6. How do we prove (disprove) S(n)? One approach is to try every value of n: is S(1) true? is S(2) true?... is S(10,000) true?... forever!!! Not very practical 6

7. Induction to the Rescue Induction is a technique for quickly proving S(n) true or false for all n we only have to do two things First show that S(1) is true do that by calculation as before continued 7

8. Second, assume that S(n) is true, and use it to show that S(n+1) is true mathematically, we show that S(n)  S(n+1) Now we know S(n) is true for all n>=1. Why? continued "  stands for "implies" 8

9. With S(1) and S(n)  S(n+1) then S(2) is true S(1)  S(2) when n == 1 With S(2) and S(n)  S(n+1) then S(3) is true S(2)  S(3) when n == 2 With S(3) and S(n)  S(n+1) then S(4) is true S(3)  S(4) when n == 3 and so on, for all n 9

10. Let’s do it Prove S(n): "n 2 + 1 > 0" for all n >= 1. First task: show S(1) is true S(1) == 1 2 + 1 == 2, which is > 0 so S(1) is true Second task: show S(n+1) is true by assuming S(n) is true continued 10

11. Assume S(n) is true, so n 2 +1 > 0 Prove S(n+1) S(n+1) == (n+1) 2 + 1 == n 2 + 2n + 1 +1 == (n 2 + 1) + 2n + 1 since n 2 + 1 > 0, then (n 2 + 1) + 2n + 1 > 0 so S(n+1) is true, by assuming S(n) is true so S(n)  S(n+1) continued 11

12. We have used induction to show two things: S(1) is true S(n)  S(n+1) is true From these it follows that S(n) is true for all n >= 1 12

13. Induction More Formally Three pieces: 1. A statement S(n) to be proved the statement must be about an integer n 2. A basis for the proof. This is the statement S(b) for some integer. Often b = 0 or b = 1. continued 13

14. 3. An inductive step for the proof. We prove the statement “S(n)  S(n+1)” for all n. The statement S(n), used in this proof, is called the inductive hypothesis We conclude that S(n) is true for all n >= b S(n) might not be true for some n < b 14

15. 3. Maths Notation Reminder Summation: means 1+2+3+4+…+n e.g.means 4+9+16+…+m 2 Product: means 1*2*3*…*n 15

16. 4. Example 1 Prove the statement S(n): for all n >= 1 e.g. 1+2+3+4 = (4*5)/2 = 10 Basis. S(1), n = 1 so 1 = (1*2)/2 continued (1) 16

17. Induction. Assume S(n) to be true. Prove S(n+1), which is: Notice that: (2) (3) continued 17

18. Substitute the right hand side (rhs) of (1) for the first operand of (3), to give: = (n 2 + n + 2n + 2) /2 = (n 2 +3n+2)/2 which is (2) 18

19. Example 2 Prove the statement S(n): for all n >= 0 e.g. 1+2+4+8 = 16-1 Basis. S(0), n = 0 so 2 0 = 2 1 -1 continued (1) 19

20. Induction. Assume S(n) to be true. Prove S(n+1), which is: Notice that: (2) (3) continued 20

21. Substitute the right hand side (rhs) of (1) for the first operand of (3), to give: = 2*(2 n+1 ) - 1 which is (2) +1 21

22. Example 3 Prove the statement S(n): n! >= 2 n-1 for all n >= 1 e.g. 5! >= 2 4, which is 120 >= 16 Basis. S(1), n = 1:1! >= 2 0 so 1 >= 1 continued (1) 22

23. Induction. Assume S(n) to be true. Prove S(n+1), which is: (n+1)! >= 2 (n+1)-1 >= 2 n (2) Notice that: (n+1)! = n! * (n+1)(3) continued 23

24. Substitute the right hand side (rhs) of (1) for the first operand of (3), to give: (n+1)! >= 2 n-1 * (n+1) >= 2 n-1 * 2 since (n+1) >= 2 (n+1)! >= 2 n which is (2) why? 24

25. Example 4 Prove the statement S(n): for all n >= 1 This proof can be used to show that. the limit of the sum: is 1 Basis. S(1), n = 1 so 1/2 = 1/2 continued (1) 25

26. Induction. Assume S(n) to be true. Prove S(n+1), which is: Notice that: (2) (3) continued 26

27. Substitute the right hand side (rhs) of (1) for the first operand of (3), to give: = which is (2) 27

28. 5. More General Inductive Proofs There can be more than one basis case. We can do a complete induction (or “strong” induction) where the proof of S(n+1) may use any of S(b), S(b+1), …, S(n) b is the lowest basis value 28

29. Example We claim that every integer >= 24 can be written as 5a+7b for non-negative integers a and b. note that some integers < 24 cannot be expressed this way (e.g. 16, 23). Let S(n) be the statement (for n >= 24) “n = 5a + 7b, for a >= 0 and b >= 0” continued 29

30. Basis. The 5 basis cases are 24 through 28. 24 = (5*2) + (7*2) 25 = (5*5) + (7*0) 26 = (5*1) + (7*3) 27 = (5*4) + (7*1) 28 = (5*0) + (7*4) continued 30

31. Induction: Let n+1 >= 29. Then n-4 >= 24, the lowest basis case. Thus S(n-4) is true, and we can write n - 4 = 5a + 7b Thus, n+1 = 5(a+1) + 7b, proving S(n+1) 31 Induction uses S(n-4)  S(n+1) Induction uses S(n-4)  S(n+1)

32. 6. Further Information Discrete Mathematics and its Applications Kenneth H. Rosen McGraw Hill, 2007, 7th edition chapter 5, section 5.1 32